Lemma 15.125.4. Let $(R,\mathfrak m)$ be a Noetherian local ring of dimension two, and let $f \in \mathfrak m$ be an element not contained in any minimal prime ideal of $R$. Then there exist $g \in \mathfrak m$ and $N \in \mathbf{N}$ such that

$f,g$ form a system of parameters for $R$.

If $h \in \mathfrak m^ N$, then $f + h, g$ is a system of parameters and $\text{length}_ R (R/(f, g)) = \text{length}_ R(R/(f + h, g))$.

**Proof.**
By Lemma 15.125.3 there exists a $g \in \mathfrak m$ such that $f, g$ is a system of parameters for $R$. Then $\mathfrak m = \sqrt{(f, g)}$. Thus there exists an $n$ such that $\mathfrak m^ n \subset (f, g)$, see Algebra, Lemma 10.32.5. We claim that $N = n + 1$ works. Namely, let $h \in \mathfrak m^ N$. By our choice of $N$ we can write $h = af + bg$ with $a, b \in \mathfrak m$. Thus

\[ (f + h, g) = (f + af + bg, g) = ((1 + a)f, g) = (f, g) \]

because $1 + a$ is a unit in $R$. This proves the equality of lengths and the fact that $f + h, g$ is a system of parameters.
$\square$

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